Integrand size = 28, antiderivative size = 28 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \left (1+x^3\right )^{3/4} \left (-3-3 x^3+7 x^4\right )}{21 x^7} \]
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Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1849, 1600, 457, 75} \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \left (x^3+1\right )^{3/4}}{3 x^3}+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^4} \]
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Rule 75
Rule 457
Rule 1600
Rule 1849
Rubi steps \begin{align*} \text {integral}& = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32 x^2-56 x^3+14 x^5-14 x^6}{x^7 \sqrt [4]{1+x^3}} \, dx \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32 x-56 x^2+14 x^4-14 x^5}{x^6 \sqrt [4]{1+x^3}} \, dx \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {32-56 x+14 x^3-14 x^4}{x^5 \sqrt [4]{1+x^3}} \, dx \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}+\frac {1}{112} \int \frac {448+112 x^3}{x^4 \sqrt [4]{1+x^3}} \, dx \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}+\frac {1}{336} \text {Subst}\left (\int \frac {448+112 x}{x^2 \sqrt [4]{1+x}} \, dx,x,x^3\right ) \\ & = \frac {4 \left (1+x^3\right )^{3/4}}{7 x^7}+\frac {4 \left (1+x^3\right )^{3/4}}{7 x^4}-\frac {4 \left (1+x^3\right )^{3/4}}{3 x^3} \\ \end{align*}
Time = 0.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \left (1+x^3\right )^{3/4} \left (-3-3 x^3+7 x^4\right )}{21 x^7} \]
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Time = 0.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89
method | result | size |
trager | \(-\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}} \left (7 x^{4}-3 x^{3}-3\right )}{21 x^{7}}\) | \(25\) |
risch | \(-\frac {4 \left (7 x^{7}-3 x^{6}+7 x^{4}-6 x^{3}-3\right )}{21 x^{7} \left (x^{3}+1\right )^{\frac {1}{4}}}\) | \(35\) |
gosper | \(-\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (7 x^{4}-3 x^{3}-3\right )}{21 x^{7} \left (x^{3}+1\right )^{\frac {1}{4}}}\) | \(36\) |
meijerg | \(\frac {4 \operatorname {hypergeom}\left (\left [-\frac {7}{3}, \frac {1}{4}\right ], \left [-\frac {4}{3}\right ], -x^{3}\right )}{7 x^{7}}+\frac {2 \sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (\frac {5 \pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {9}{4}\right ], \left [2, 3\right ], -x^{3}\right )}{32 \Gamma \left (\frac {3}{4}\right )}-\frac {\left (3-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{4 \Gamma \left (\frac {3}{4}\right )}-\frac {\pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right ) x^{3}}\right )}{3 \pi }+\frac {5 \operatorname {hypergeom}\left (\left [-\frac {4}{3}, \frac {1}{4}\right ], \left [-\frac {1}{3}\right ], -x^{3}\right )}{4 x^{4}}+\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {\pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], -x^{3}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{6 \pi }+\frac {\operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {1}{4}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}\) | \(180\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \, {\left (7 \, x^{4} - 3 \, x^{3} - 3\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{21 \, x^{7}} \]
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Result contains complex when optimal does not.
Time = 2.51 (sec) , antiderivative size = 177, normalized size of antiderivative = 6.32 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=- \frac {\Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} - \frac {5 \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{4} \\ - \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{4} \Gamma \left (- \frac {1}{3}\right )} - \frac {4 \Gamma \left (- \frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{3}, \frac {1}{4} \\ - \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{7} \Gamma \left (- \frac {4}{3}\right )} - \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} - \frac {4 \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {15}{4}} \Gamma \left (\frac {9}{4}\right )} \]
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Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \, {\left (7 \, x^{7} - 3 \, x^{6} + 7 \, x^{4} - 6 \, x^{3} - 3\right )}}{21 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{4}} {\left (x + 1\right )}^{\frac {1}{4}} x^{7}} \]
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\[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=\int { \frac {{\left (x^{4} - x^{3} - 1\right )} {\left (x^{3} + 4\right )}}{{\left (x^{3} + 1\right )}^{\frac {1}{4}} x^{8}} \,d x } \]
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Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=\frac {12\,{\left (x^3+1\right )}^{3/4}+12\,x^3\,{\left (x^3+1\right )}^{3/4}-28\,x^4\,{\left (x^3+1\right )}^{3/4}}{21\,x^7} \]
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